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Algebraic functions

Reciprocal Function f(x) = 1/x: Graph, Derivative, Integral and Limits

The reciprocal function 1/x: its hyperbola graph and asymptotes, the derivative -1/x², why its integral is ln|x| + C, and its one-sided limits.

Algebraic functions Updated 3 min read

At a glance

f(x) = \frac{1}{x}
Domain
x \ne 0
Range
y \ne 0
Derivative
-\frac{1}{x^2}
Integral
\ln|x| + C
Asymptotes
x = 0 and y = 0
Symmetry
Odd: f(-x) = -f(x)
On this page
  1. Formula
  2. Graph of 1/x
  3. Key properties
  4. Limits of 1/x
  5. Derivative of 1/x
  6. Integral of 1/x
  7. Worked examples
  8. Common mistakes
  9. How it connects

The reciprocal function f(x) = \frac{1}{x} sends each number to one over itself. It is the simplest rational function, the power function with exponent -1, and the standard example of infinite limits, asymptotes and the natural logarithm's derivative.

Formula

f(x) = \frac{1}{x} = x^{-1}, \qquad x \ne 0

Graph of 1/x

The graph is a hyperbola with two branches, one in the first quadrant and one in the third.

Graph of the reciprocal function y = 1/x
The graph approaches the vertical asymptote x = 0 and the horizontal asymptote y = 0 without touching either.

Key properties

  • Domain and range. All reals except 0, for both.
  • Odd function. \frac{1}{-x} = -\frac{1}{x}, so the graph is symmetric about the origin.
  • Asymptotes. Vertical asymptote x = 0, horizontal asymptote y = 0.
  • Its own inverse. Solving y = \frac{1}{x} for x gives x = \frac{1}{y}, so f^{-1} = f and the graph is symmetric about y = x.
  • Discontinuous only at 0, which is not in the domain. It is continuous on its domain.

Limits of 1/x

\lim_{x \to 0^+}\frac{1}{x} = +\infty, \qquad \lim_{x \to 0^-}\frac{1}{x} = -\infty, \qquad \lim_{x \to \pm\infty}\frac{1}{x} = 0

Because the one-sided limits at 0 disagree, \lim_{x \to 0}\frac{1}{x} does not exist. The limit at infinity is the basic building block for limits at infinity of every rational function.

Derivative of 1/x

\frac{d}{dx}\,\frac{1}{x} = -\frac{1}{x^2}

This follows from the power rule with n = -1. The derivative is negative everywhere it is defined, so the function decreases on each branch, and the tangent lines get steeper near 0 and flatter far away. The second derivative is \frac{2}{x^3}: concave up for x > 0, concave down for x < 0.

Integral of 1/x

\int \frac{1}{x}\,dx = \ln|x| + C

The reverse power rule would give \frac{x^0}{0}, which is undefined, so \frac{1}{x} is the one power that integrates to a natural logarithm. The absolute value covers negative x: for x < 0, \frac{d}{dx}\ln(-x) = \frac{-1}{-x} = \frac{1}{x}.

A definite integral cannot cross x = 0: \int_{-1}^{1}\frac{1}{x}\,dx is an improper integral that diverges.

Worked examples

Example 1: tangent line. At x = 2, f(2) = \frac{1}{2} and f'(2) = -\frac{1}{4}, so the tangent line is y = \frac{1}{2} - \frac{1}{4}(x - 2) = -\frac{1}{4}x + 1.

Example 2: definite integral. \int_1^e \frac{1}{x}\,dx = \ln e - \ln 1 = 1.

Example 3: a shifted reciprocal. \frac{d}{dx}\frac{3}{x - 1} = -\frac{3}{(x - 1)^2} and \int \frac{3}{x - 1}\,dx = 3\ln|x - 1| + C.

Common mistakes

  1. Writing \int \frac{1}{x}\,dx = \frac{x^0}{0} or \ln x without the absolute value.
  2. Getting the derivative sign wrong. \frac{d}{dx}\frac{1}{x} is -\frac{1}{x^2}, not \frac{1}{x^2}.
  3. Confusing \frac{1}{x} with x^{-1} as an inverse function. \frac{1}{f(x)} is a reciprocal, while f^{-1}(x) usually means an inverse. For \frac{1}{x} they happen to agree, which is unusual.
  4. Integrating across the asymptote as if the function were continuous there.

How it connects

The area under \frac{1}{t} from 1 to x is exactly \ln x, which is one way to define the natural logarithm. The power functions page covers the family x^n that \frac{1}{x} belongs to, and the harmonic series \sum \frac{1}{n} is its discrete cousin.

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One-sided and infinite limits

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Frequently asked questions

What is the derivative of 1/x?

Write \frac{1}{x} = x^{-1} and use the power rule: \frac{d}{dx}x^{-1} = -x^{-2} = -\frac{1}{x^2}.

What is the integral of 1/x?

The integral is \int \frac{1}{x}\,dx = \ln|x| + C. The reverse power rule fails here because it would divide by n + 1 = 0. The absolute value makes the formula valid for negative x too.

What is the limit of 1/x as x approaches 0?

The limit does not exist. From the right \frac{1}{x} \to +\infty and from the left \frac{1}{x} \to -\infty, so the two one-sided limits disagree.

Is 1/x decreasing?

It decreases on (-\infty, 0) and on (0, \infty) separately, since its derivative is negative. It is not decreasing on its whole domain, because \frac{1}{-1} = -1 < 1 = \frac{1}{1}.

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